.)Specific atomic models:
J.J. Thomson, now well respected, tried to model atoms as a lump of positive
matter, with electrons embedded inside like raisins in a fruitcake. (The "plum
pudding model") So, if the radius of hydrogen is about 10^-10m, he pictured
this as a smeared out positive charge, with a tiny electron
inside. (Tiny, because based on electromagnetic theory, a fundamental ball of
charge, with radius r, has E.M. energy = e^2/r. If this energy is responsible
for the mass of the electron, then
.)
Now, people at this time know of spectral lines, the radiation emitted by atoms. Perhaps this was caused by the electrons wiggling in the pudding? Zeeman studied the effect of B fields on spectral lines, and found they had q/m = e/m = 2000 times more then q(H)/m(H), i.e., that whatever was radiating inside the hydrogen was the same as Thomsons electron! So, atoms have these "cathode ray electrons" inside them, its not just something fabricated in high voltage electron guns.
Thomson's model gives a very natural oscillation frequency:
The electron feels a net inward force, because it's sitting in a cloud of + charge. The force is given by
This is pure SHM,
(Energy = 10 eV, the first hydrogen "Lyman" line is 10.2 eV)
Not such bad agreement with experiment. But, how do you get other frequencies out? (This model only gives one vibration mode) Is this model really correct? Rutherford said, the best way to find out what's inside a fruitcake is to put your finger in it:
Rutherford Scattering
Geiger and Marsden, in 1911, used fast alpha particles (which are really He++ nuclei, i.e. 2 protons and 2 neutrons bound together in a tiny object).
(This screen flashes when struck by alphas). The expectation was that the alphas should slam right on through the foil. In Thomson's model, the electric charge in an atom is uniformly spread out over an area of size about 1 Angstrom.
With this charge density, one predicts deflections of 1 degree or less.
Geiger and Marsden saw that, although most of the alphas did go through as expected, some scattered through very large angles. Rutherford: "It was as incredible as if you fired a 15 inch shell at a piece of tissue paper, and it came back and hit you".
Alpha's weigh about 8000 * m_e, and in this experiment had a velocity of about 2*10^7 m/s. So, they're very heavy and very fast! To be deflected, there must be some very strong forces acting on them. Rutherford proposed that the atom had a tiny, positively charged nucleus containing (nearly) all of the mass, and the electron is farther away. The atom is thus mostly empty space, and most alphas go right through. But, if an alpha happens to go near a nucleus, the intense E field scatters it a lot. (Electrons don't really have much to do with the scattering, because they're so light)
Comparing foils of different metal showed that different nuclei have different charges, and give different amounts of deflection. This was a way to measure the charge of an atom's nucleus. Result: Each element has a unique charge, +Ze (== atomic number), which we now know to come from Z protons in the nucleus.
Some numbers:
If Ze for gold is spread out over an Angstrom or so (a la Thomson), then the
maximum E field is at the surface,
.
If instead you estimate the E field at the surface of a gold nucleus (with a modern idea of the size of the nucleus), the radius is 10,000 times smaller, so r^2 is 10^8 times smaller, i.e. E is 10^8 times bigger, or 10^21 V/m. You need this kind of E field to get the large deflections seen by Rutherford and his students.
Rutherford's model was accepted primarily because he used it to derive a formula for the number of alphas scattering to a given angle, which agreed well with the experiment. The idea is as follows:
If the nucleus is heavy, and sits still (a decent approximation), then the scattering is elastic, and the momentum and energy of the alpha don't change in magnitude (only the direction of p changes)
.
(Can you see why from the picture?)
If we knew the impact parameter, "b", then we could figure out theta.
After all, Newton's law says F=(dp/dt) so
.
And, since v is constant, t just sweeps uniformly along the hyperbole I drew,
and F=(2e)(Ze)/r^2 is known at each point along the way, so this just becomes a
(slightly tricky) line integral. The result (which takes a couple of lines of
math and some basic classical mechanics) is
,
which we can rewrite as
Even without working it out, this should at least look plausible:
Large mass (or velocity) -> small angle theta (sensible)
Large b (or small Z) -> small angle theta (sensible)
Small b means bigger forces -> larger angle theta (sensible)
If you solve for the scattering angle, you get
0.5*
.
But, with a beam of alphas on a foil target, we won't ever really know what "b" is for an individual collision - it'll be just about anything. So, let's try to eliminate the unknown "b" from our discussion.
Suppose I = intensity of alphas = # of alphas/sec/unit area.
Let's also suppose our target was only one atom (we'll generalize soon) Those alphas that come in with impact parameter smaller than b will scatter at an angle more than the theta calculated above.
In fact, any alpha coming in inside this circle of radius "b" around the atom will scatter out beyond angle theta.
This means that
(the # of alphas scattered beyond
)/sec
=
[the right side is the (# incident/sec/area * area)]
Now generalize this to a real foil:
(# alphas observed beyond
)/sec
=
* (total # of atoms struck)
=
*(# atoms/unit volume)*(Volume of foil being struck)
=
where A = area of beam, and t=thickness of foil, and M is the gram formula weight of the target material = the number of grams/mole.
Finally, the fraction of alphas scattered beyond theta is this last equation divided by the total number of alphas hitting the foil,
(where I used my earlier result to eliminate b)
Everything here is experimentally measurable. Geiger and Marsden counted 100,000 flashes, and confirmed this equation.
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